Consider the following system of linear equations:x-y+2z=1,2x+y+3z=3,x+3y+z=2.What is the solution to this system?

To solve the system of equations, we can use the method of substitution or elimination. Here, we will use the method of elimination.

The system of equations is:

1. x - y + 2z = 1
2. 2x + y + 3z = 3
3. x + 3y + z = 2

First, we can eliminate y from the first two equations. To do this, we add the first and the second equations:

(1) + (2) => x - y + 2z + 2x + y + 3z = 1 + 3 => 3x + 5z = 4

Now, we have a new system of equations:

4. 3x + 5z = 4
5. x + 3y + z = 2

Next, we can eliminate x from equations (4) and (5). To do this, we multiply the second equation by 3 and subtract the first equation from the result:

3*(5) - (4) => 3x + 9y + 3z - 3x - 5z = 6 - 4 => 9y - 2z = 2

Now, we have a new system of equations:

6. 9y - 2z = 2
7. 3x + 5z = 4

To find the value of y, we can rearrange equation (6):

9y = 2z + 2 => y = (2z + 2) / 9

Substitute y in equation (5):

x + 3((2z + 2) / 9) + z = 2 => x + 2z/3 + 2/3 + z = 2 => x + 5z/3 = 2 - 2/3 => x = 2/3 - 5z/3

Now we have:

8. x = 2/3 - 5z/3
9. y = (2z + 2) / 9

Substitute x and y in equation (1):

(2/3 - 5z/3) - ((2z + 2) / 9) + 2z = 1 => 2/3 - 5z/3 - 2z/9 - 2/9 + 2z = 1 => 2/3 - 15z/9 - 2z/9 - 2/9 + 18z/9 = 1 => 2/3 - 2/9 = 1z/9 => 4/9 = z/9 => z = 4

Substitute z = 4 in equations (8) and (9) to find x and y:

x = 2/3 - 54/3 = -6 y = (24 + 2) / 9 = 1

So, the solution to the system of equations is x = -6, y = 1, z = 4.